Jim Randell

Updates from Jim Randell Toggle Comment Threads | Keyboard Shortcuts

Jim Randell 9:49 am on 25 February 2025 Permalink | Reply
Tags: by: Bob Walker ( 10 )
Teaser 2504: [Hidden crosses]
From The Sunday Times, 19th September 2010 [link] [link]

Here are 18 black cards and there is a cross on the back of some of them.

The clue in any gap indicates how many crosses there are on cards that are adjacent to that gap.

How many black cards have a cross on them?

This puzzle was originally published with no title.

[teaser2504]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
ruudvanderham and Jim Randell are discussing. Toggle Comments
- Jim Randell 9:50 am on 25 February 2025 Permalink | Reply
  Here is a solution using the [[ SubstitutedExpression ]] solver from the enigma.py library.
  
  It runs in 80ms. (And the internal runtime of the generated program is just 53µs).
```
#! python3 -m enigma -rr

SubstitutedExpression

#  +---+---+---+---+---+---+
#  | A |   | B |   | C |!=0|
#  +---+---+---+---+---+---+
#  | 1 | D | >2| E | <2| F |
#  +---+---+---+---+---+---+
#  | G |   | H |!=1| I |   |
#  +---+---+---+---+---+---+
#  |   | J |   | K | >1| L |
#  +---+---+---+---+---+---+
#  | M | >2| N |!=3| P |   |
#  +---+---+---+---+---+---+
#  |!=2| Q |   | R | =1| S |
#  +---+---+---+---+---+---+

--distinct=""
--digits="0,1"

# the clues
"C + F != 0"
"A + D + G = 1"
"B + D + E + H > 2"
"C + E + F + I < 2"
"E + H + I + K != 1"
"I + K + L + P > 1"
"J + M + N + Q > 2"
"K + N + P + R != 3"
"M + Q != 2"
"P + R + S = 1"

--answer="sum([A, B, C, D, E, F, G, H, I, J, K, L, M, N, P, Q, R, S])"
--template=""
```
  Solution: 10 of the black cards have a cross on them.
  
  There are 4 possible arrangements, that can be summarised as follows:
  
  8 of the crosses are marked X, and the remaining 2 crosses appear in squares marked a and b.
  
  LikeLike
- ruudvanderham 12:26 pm on 1 March 2025 Permalink | Reply
```
import itertools


def s(*args):
    return sum(a[arg] for arg in args)


for a in itertools.product((0, 1), repeat=18):
    if (
        s(2, 5) != 0
        and s(0, 3, 6) == 1
        and s(1, 3, 4, 7) > 2
        and s(2, 4, 5, 8) < 2
        and s(4, 7, 8, 10) != 1
        and s(8, 10, 11, 14) > 1
        and s(9, 12, 13, 15) > 2
        and s(10, 13, 14, 16) != 3
        and s(12, 15) != 2
        and s(14, 16, 17) == 1
    ):
        print(a, sum(a))
```
  LikeLike
Reply Cancel reply
Required fields are marked *

Name *

Email *

Website

Notify me of new comments via email.
Notify me of new posts via email.
Δ
Jim Randell 6:42 am on 23 February 2025 Permalink | Reply
Tags: by: John Fletcher ( 2 )
Teaser 3257: Powers of deduction
From The Sunday Times, 23rd February 2025 [link] [link]

One evening, Sherlock Holmes challenges Watson and Mycroft to a game of Cluedo.

He places one each of the six “person” cards, six “weapon” cards and nine “room” cards into the murder bag, shuffles the rest and deals them out equally.

Mycroft sees his cards and says, “If I guessed what was in the murder bag now, my chance of being correct is one in an integer that is a power greater than two”.

“I’m saying nothing”. Watson replies. “I’d only give something away”.

“Commendably prudent”, Sherlock says, “but the same is true for you”.

Watson studies his cards, then looks agitated. “How could you know that?”

“Sherlock isn’t cheating”. Mycroft reassures him. “He didn’t even know how many person cards you’re holding”.

How many room cards is Mycroft holding?

[teaser3257]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 8:14 am on 23 February 2025 Permalink | Reply
  After the culprit is selected there are 5 person, 5 weapon and 8 room cards remaining, and these are distributed among the 3 players (so, 6 cards each).
  
  If a player is holding p person cards, w weapon cards, and r room cards, then the probability of making a correct accusation from the cards he is not holding is:
  
  P = (1/(6 − p)) × (1/(6 − w)) × (1/(9 − r))
  1/P = (6 − p) × (6 − w) × (9 − r)
  
  This Python program finds possible (p, w, r) hands that have a 1/(n^k) chance of giving a correct guess, where k > 2.
```
from enigma import (decompose, irange, inf, powers, first, lt, printf)

# there are only a few higher powers than squares to consider
(M, pows) = (6 * 6 * 9, set())
for k in irange(3, inf):
  ps = first(powers(2, inf, k), count=lt(M))
  if not ps: break
  pows.update(ps)

# find hands of 6 cards, with a 1/power chance of a correct guess
for (p, w, r) in decompose(6, 3, increasing=0, sep=0, min_v=0):
  if p > 5 or w > 5: continue
  N = (6 - p) * (6 - w) * (9 - r)
  if N not in pows: continue
  # output possible hands
  printf("p={p} w={w} r={r} -> N={N}")
```
  The only cards that give a required 1/power chance are:
  
  (p w r) = (3 3 0) = 1/81 (81 = 3^4)
  (p w r) = (1 1 4) = 1/125 (125 = 5^3)
  
  Mycroft declares he has one of these hands, and Sherlock declares Watson has one of these hands. It is not possible for there to be more than one (3 3 0) hand, but if there are a (3 3 0) and a (1 1 4) then the remaining hand is also a (1 1 4), and if there are two (1 1 4) hands then the remaining hand is a (3 3 0), and so Sherlock also has one of these hands.
  
  Hence the three hands are (1 1 4) (1 1 4) (3 3 0) (in some order). So when Mycroft declares he has one of these, Sherlock can see that he (Sherlock) also has one, and so Sherlock can declare that Watson must too have one of these hands.
  
  Which means Mycroft and Sherlock both know that each of them has one of these three hands, and if either of them has (3 3 0) they know the other two must both have (1 1 4).
  
  So Sherlock must be holding (1 1 4) (as Mycroft states that he does not know which of the possible hands Watson is holding), and the only way Mycroft can be certain that Sherlock has (1 1 4) is if he (Mycroft) is holding (3 3 0).
  
  And so: Mycroft = (3 3 0), Watson = (1 1 4), Sherlock = (1 1 4).
  
  Solution: Mycroft is holding no room cards.
  
  LikeLike
  - Jim Randell 9:44 pm on 23 February 2025 Permalink | Reply
    I’ve not seen another solution to this puzzle yet, so here is a complete solution.
    
    The following Python program runs in 60ms. (Internal runtime is 476µs).
```
from enigma import (
  namedtuple, decompose, irange, inf, powers, first, lt,
  subsets, filter_unique, unpack, intersect, seq2str, printf
)

# there are only a few higher powers than squares to consider
(M, pows) = (6 * 6 * 9, set())
for k in irange(3, inf):
  ps = first(powers(2, inf, k), count=lt(M))
  if not ps: break
  pows.update(ps)

# find hands of 6 cards, with a 1/power chance of a correct guess
Hand = namedtuple('Hand', 'p w r')
hs = list()
for (p, w, r) in decompose(6, 3, increasing=0, sep=0, min_v=0):
  if p > 5 or w > 5: continue
  N = (6 - p) * (6 - w) * (9 - r)
  if N not in pows: continue
  # output possible hands
  printf("[({p}, {w}, {r}) -> N={N}]")
  hs.append(Hand(p, w, r))

# collect possible scenarios
ss = list()
# M and W each have one of these hands
for (M, W) in subsets(hs, size=2, select='M'):
  # and S has the remaining cards
  S = Hand(5 - M.p - W.p, 5 - M.w - W.w, 8 - M.r - W.r)
  if S.p < 0 or S.w < 0 or S.r < 0: continue
  ss.append((M, W, S))

# find situations where S _cannot_ determine W.p
ss1 = filter_unique(ss, unpack(lambda M, W, S: S), unpack(lambda M, W, S: W.p)).non_unique

# find situations where M _can_ fully determine S's hand
ss2 = filter_unique(ss, unpack(lambda M, W, S: M), unpack(lambda M, W, S: S)).unique

# look for common situations
for (M, W, S) in intersect([ss1, ss2]):
  printf("M={M} W={W} S={S}", M=seq2str(M), W=seq2str(W), S=seq2str(S))
```
    LikeLike
Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply
Tags: by: Victor Bryant ( 148 )
Teaser 2585: Easter Teaser
From The Sunday Times, 8th April 2012 [link] [link]

I took the letters A, E, E, R, S and T and wrote down the long list of all possible different combinations of them. So, in alphabetical order, my list was AEERST, AERTS, …, EASTER, …, TEASER, …, TSREEA. I then assigned each of the letters a different odd digit, turning my list into a list of numbers. Surprisingly, the grand total of my list of numbers contained no odd digit.

What is E+A+S+T+E+R?

This completes the archive of Teaser puzzles from 2012, so there is now a complete archive of puzzles (and solutions) from Teaser 2569 (December 2011) to the most recent puzzle Teaser 3256 (February 2025), along with many earlier puzzles. There are another 21 puzzles to post from 2011 before all puzzles from the 2020 book are posted.

Between S2T2 and Enigmatic Code there are now 3577 puzzles available.

[teaser2585]
Rate this:
Share:
X
Facebook
WhatsApp
Email
Like Loading...
Jim Randell is discussing. Toggle Comments
- Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply
  The following Python runs in 59ms. (Internal runtime is 367µs).
```
from enigma import (irange, multiset, repdigit, subsets, nsplit, map2str, printf)

word = 'EASTER'
m = multiset.from_seq(word)

# calculate multiplier for sum of letters
n = m.size()
p = repdigit(n) * m.multinomial() // n

# even/odd digits
evens = set(irange(0, 9, step=2))
odds = irange(1, 9, step=2)

# now assign odd digits to each letter
rs = set()
ks = sorted(m.keys())
for vs in subsets(odds, size=len(ks), select='P'):
  d = dict(zip(ks, vs))
  # calculate the sum of the letters in the word
  s = sum(d[x] for x in word)
  # calculate the sum of all the permutations
  t = s * p
  # check the total contains only even digits
  if not evens.issuperset(nsplit(t)): continue
  # answer is sum of the letters in word
  printf("[sum = {s}; {d} -> {t}]", d=map2str(d))
  rs.add(s)

printf("answer = {rs}", r=sorted(rs))
```
  Solution: E+A+S+T+E+R = 34.
  
  E is 9, and A, R, S, T are 1, 3, 5, 7 in some order.
  
  And the sum of all possible arrangements is 226666440.
  
  For EASTER there are 360 different arrangements of the letters (there are 2 E‘s that are not distinguishable).
  
  There are 6 possible positions each letter can appear in. And each of the 6 letters (including repeats) appears in each position 60 times.
  
  So each letter contributes 60 × 111111 = 6666660 times its own value to the total sum.
  
  With the sum of the letters in EASTER = 1 + 3 + 5 + 7 + 2×9 = 34 we arrive at a total sum of:
  
  total = 34 × 6666660 = 226666440
  
  which consists entirely of even digits.
  
  And we see that reordering the values of the non-repeated letters does not change the total.
  
  Using two copies of any odd digit other than 9 gives a total with at least one odd digit.
  
  LikeLike

Jim Randell 8:30 am on 18 February 2025 Permalink | Reply
Tags: by: Neil MacKinnon

Teaser 3256: Domino-do

From The Sunday Times, 16th February 2025 [link] [link]

I have a standard set of 28 dominoes with a number of spots from zero to six at each end, and each possible pair of numbers occurring once in the set. When two dominoes in a line touch, the two adjacent ends must have the same number of spots.

I started a line by placing one of the dominoes on a table. Then, in correct domino style, I placed another adjacent to it at the right-hand end. I continued in this way (placing each domino at either end of the line) until more than 14 dominoes were lined up. At each stage the total number of spots on the table was a different odd number less than 100. In fact, the last two totals were prime but the first two totals were not.

From the left, what were the first six dominoes in the line (in the form 1-4, 4-0, …)?

[teaser3256]

Jim Randell 8:25 am on 16 February 2025 Permalink | Reply

The totals are all (different) odd numbers, so we must start with an odd domino (that is not prime), and each subsequently placed domino must be even (and not [0-0]). So at most one of the other even dominoes can remain unused.

Here is a constructive program. It starts by find possible placements for the first and second dominoes (an odd domino, and then a non-zero even domino, such that the totals are both non-prime), and then considers viable sets of at least 13 more even dominoes, that give a prime total for the entire line, and then looks for the final placement (that also gives a prime total after the penultimate placement), and then tries to construct a sequence of plays that gives a viable line of dominoes.

This Python 3 program runs in 625ms.

from enigma import (irange, subsets, is_prime, filter2, rev, remove, unpack, delete, multiset, printf)

# a standard set of dominoes
ds = list(subsets(irange(0, 6), size=2, select='R'))
ds.remove((0, 0))  # but we can't play [0-0], as totals must be all different

# canonical form of a domino
def canonical(d): (x, y) = d; return (d if x < y else (y, x))

# add <k> dominoes from <ds> to the line <ss> (totals so far = <ts>)
# return (<domino line>, <totals>, <unused dominoes>)
def solve(k, ds, ss, ts):
  # are we done?
  if not k:
    yield (ss, ts, ds)
  else:
    # find candidates for the next domino
    ps = list()
    for (x, y) in ds:
      # place right
      if ss[-1][-1] == x:
        ps.append((-1, (x, y)))
      if x != y and ss[-1][-1] == y:
        ps.append((-1, (y, x)))
      # place left
      if ss[0][0] == x:
        ps.append((0, (y, x)))
      if x != y and ss[0][0] == y:
        ps.append((0, (x, y)))
    # make the placement
    for (i, d) in ps:
      t = ts[-1] + sum(d)
      ds_ = remove(ds, canonical(d))
      ss_ = (ss + [d] if i == -1 else [d] + ss)
      yield from solve(k - 1, ds_, ss_, ts + [t])

# split the dominoes into odd and even
(odds, evens) = filter2((lambda x: sum(x) % 2 == 1), ds)

# collect required answers
rs = multiset()

# we start with an odd (non-prime) domino
for d in odds:
  t = sum(d)
  if is_prime(t): continue

  # find a placement for the second (even) domino
  for (ss2, ts2, ds2) in solve(1, evens, [d], [t]):
    # which gives a non-prime total
    if is_prime(ts2[-1]): continue
    # the first placement is on the right, so reflect if necessary
    if canonical(ss2[1]) == d: ss2 = rev(rev(x) for x in ss2)
    # count the number of times each end occurs
    m = multiset.from_seq(*ss2)

    # look for a set of at least 13 additional even dominoes to place
    for ds in subsets(ds2, min_size=13):
      # calculate total number of pips (must be a prime, < 100)
      t = ts2[-1] + sum(sum(x) for x in ds)
      if not (t < 100 and is_prime(t)): continue
      # in the complete line at most 2 ends can appear an odd number of times
      if sum(n % 2 for n in m.combine(*ds).values()) > 2: continue

      # consider the final domino, so the penultimate total is prime
      for (i, dx) in enumerate(ds):
        tx = t - sum(dx)
        if not is_prime(tx): continue

        # place all the other dominoes
        printf("[trying {ss2} -> {dx}]")
        dsx = delete(ds, [i])
        for (ssx, tsx, _) in solve(len(dsx), dsx, ss2, ts2):
          # can we place the final domino?
          for (ss, ts, _) in solve(1, [dx], ssx, tsx):
            printf("[{d} -> {ss} {ts}]")
            rs.add(tuple(ss[:6]))

# output the answer(s)
for (ss, n) in rs.most_common():
  printf("{ss} ... [{n} solutions]")

Solution: The leftmost dominoes are: [5-3] [3-3] [3-1] [1-1] [1-5] [5-5].

There are two possibilities for the completed line of dominos, each uses the same set of 15 dominos:

[5-3] [3-3] [3-1] [1-1] [1-5] [5-5] [5-4] [4-2] [2-2] [2-0] [0-4] [4-4] [4-6] [6-6] [6-0]
[5-3] [3-3] [3-1] [1-1] [1-5] [5-5] [5-4] [4-2] [2-2] [2-0] [0-6] [6-6] [6-4] [4-4] [4-0]

The lines are the same for the leftmost 10 dominoes, and then the rightmost 5 dominoes may appear in either direction.

The first play is the odd domino [5-4] (total = 9; odd, not prime), and this is followed on the right by [4-2] (total = 15; odd, not prime).

The remaining dominoes are played (in any viable order) until there is only [5-3] remaining to place on the left end. Before it is played the total is 89 (odd, prime), and after it is played the total is 97 (odd, prime).

The unused even domino is [2-6].

An even domino must have ends that are the same parity, and so since we start with an odd domino (which must have ends that are of different parities) the even numbers accumulate on the even side of the initial domino, and the odd numbers accumulate on the odd side of the initial domino. And when the line is complete every end apart from the two outside ends must be paired with its neighbour, and so must appear an even number of times. The exceptions are the outside ends of the line, so there must be two ends (one of them even, and one of them odd) that appear an odd number of times. And this is the basis for my faster program below that just looks for viable final arrangements of dominoes.

Search

Recent Posts

Recent Comments

Archives

Categories

Subscribe

RSS Links

Site Stats

Meta

Updates from Jim Randell Toggle Comment Threads | Keyboard Shortcuts

Jim Randell 9:49 am on 25 February 2025 Permalink | Reply Tags: by: Bob Walker ( 10 )

Rate this:

Share:

Jim Randell 9:50 am on 25 February 2025 Permalink | Reply

ruudvanderham 12:26 pm on 1 March 2025 Permalink | Reply

Reply Cancel reply

Jim Randell 6:42 am on 23 February 2025 Permalink | Reply Tags: by: John Fletcher ( 2 )

Rate this:

Share:

Jim Randell 8:14 am on 23 February 2025 Permalink | Reply

Jim Randell 9:44 pm on 23 February 2025 Permalink | Reply

Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply Tags: by: Victor Bryant ( 148 )

Rate this:

Share:

Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply

Jim Randell 8:30 am on 18 February 2025 Permalink | Reply Tags: by: Neil MacKinnon

Rate this:

Share:

Jim Randell 8:30 am on 18 February 2025 Permalink | Reply

Frits 10:33 am on 22 February 2025 Permalink | Reply

Frits 3:02 am on 23 February 2025 Permalink | Reply

Jim Randell 7:14 am on 16 February 2025 Permalink | Reply Tags: by: Victor Bryant ( 148 )

Rate this:

Share:

Jim Randell 8:25 am on 16 February 2025 Permalink | Reply

Jim Randell 10:52 am on 18 February 2025 Permalink | Reply

Frits 6:33 pm on 17 February 2025 Permalink | Reply

Frits 1:15 pm on 20 February 2025 Permalink | Reply

Jim Randell 9:34 am on 14 February 2025 Permalink | Reply Tags: by: Victor Bryant ( 148 )

Rate this:

Share:

Jim Randell 9:34 am on 14 February 2025 Permalink | Reply

Jim Randell 9:21 am on 11 February 2025 Permalink | Reply Tags: by: Roger Webster

Rate this:

Share:

Jim Randell 9:21 am on 11 February 2025 Permalink | Reply

Jim Randell 7:47 am on 9 February 2025 Permalink | Reply Tags: by: Mark Valentine ( 17 )

Rate this:

Share:

Jim Randell 7:59 am on 9 February 2025 Permalink | Reply

Jim Randell 1:31 pm on 9 February 2025 Permalink | Reply

Frits 8:29 am on 9 February 2025 Permalink | Reply

Jim Randell 8:16 am on 7 February 2025 Permalink | Reply Tags: by: Andrew Skidmore ( 88 )

Rate this:

Share:

Jim Randell 8:16 am on 7 February 2025 Permalink | Reply

Frits 4:07 pm on 7 February 2025 Permalink | Reply

John Crabtree 4:58 am on 14 February 2025 Permalink | Reply

Jim Randell 9:15 am on 4 February 2025 Permalink | Reply Tags: by: Angela Newing ( 40 )

Rate this:

Share:

Jim Randell 9:17 am on 4 February 2025 Permalink | Reply

Jim Randell 7:02 am on 2 February 2025 Permalink | Reply Tags: by: Peter Good ( 29 )

Rate this:

Share:

Jim Randell 7:22 am on 2 February 2025 Permalink | Reply

Frits 6:36 am on 3 February 2025 Permalink | Reply

Jim Randell 10:39 am on 31 January 2025 Permalink | Reply Tags: by: C Higgins ( 39 ), by: H Bradley ( 39 )

Rate this:

Share:

Jim Randell 10:40 am on 31 January 2025 Permalink | Reply

Ruud 3:58 pm on 2 February 2025 Permalink | Reply

GeoffR 9:03 am on 1 February 2025 Permalink | Reply

Jim Randell 10:41 am on 28 January 2025 Permalink | Reply Tags: by: Angela Newing ( 40 )

Rate this:

Share:

Jim Randell 10:42 am on 28 January 2025 Permalink | Reply

Jim Randell 1:45 am on 26 January 2025 Permalink | Reply Tags: by: Colin Vout ( 39 )

Rate this:

Share:

Jim Randell 9:49 am on 25 February 2025 Permalink | Reply
Tags: by: Bob Walker ( 10 )

Jim Randell 6:42 am on 23 February 2025 Permalink | Reply
Tags: by: John Fletcher ( 2 )

Jim Randell 12:09 pm on 21 February 2025 Permalink | Reply
Tags: by: Victor Bryant ( 148 )

Jim Randell 8:30 am on 18 February 2025 Permalink | Reply
Tags: by: Neil MacKinnon

Jim Randell 7:14 am on 16 February 2025 Permalink | Reply
Tags: by: Victor Bryant ( 148 )

Jim Randell 9:34 am on 14 February 2025 Permalink | Reply
Tags: by: Victor Bryant ( 148 )

Jim Randell 9:21 am on 11 February 2025 Permalink | Reply
Tags: by: Roger Webster

Jim Randell 7:47 am on 9 February 2025 Permalink | Reply
Tags: by: Mark Valentine ( 17 )

Jim Randell 8:16 am on 7 February 2025 Permalink | Reply
Tags: by: Andrew Skidmore ( 88 )

Jim Randell 9:15 am on 4 February 2025 Permalink | Reply
Tags: by: Angela Newing ( 40 )

Jim Randell 7:02 am on 2 February 2025 Permalink | Reply
Tags: by: Peter Good ( 29 )

Jim Randell 10:39 am on 31 January 2025 Permalink | Reply
Tags: by: C Higgins ( 39 ), by: H Bradley ( 39 )

Jim Randell 10:41 am on 28 January 2025 Permalink | Reply
Tags: by: Angela Newing ( 40 )

Jim Randell 1:45 am on 26 January 2025 Permalink | Reply
Tags: by: Colin Vout ( 39 )

Jim Randell 7:58 am on 24 January 2025 Permalink | Reply
Tags: by: Graham Smithers ( 86 )

Jim Randell 7:49 am on 21 January 2025 Permalink | Reply
Tags: by: Danny Roth ( 88 )

Jim Randell 6:58 am on 19 January 2025 Permalink | Reply
Tags: by: Danny Roth ( 88 )

Jim Randell 7:42 am on 17 January 2025 Permalink | Reply
Tags: by: C Higgins ( 39 ), by: H Bradley ( 39 )

Jim Randell 10:07 am on 14 January 2025 Permalink | Reply
Tags: by: Graham Smith

Jim Randell 7:00 am on 12 January 2025 Permalink | Reply
Tags: by: Howard Williams ( 45 )